线性响应理论（Linear Response Theory）

多体理论

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Retarded GF, i.e., $G^R(t)$ describes the reaction of the system to external perturbation. Up to now, we can only calculate the GF in equilibrium, which means no external perturbation.

But, we still can use the equilibrium GF to calculate the linear reaction of the system in a first order perturbation. This is so-called linear response theory.

Remember, Hooke's law,

$$\Delta x = \frac{1}{k} F$$

and Ohm's law,

$$I = \frac{1}{R}V$$

where $F$ and $V$ are external perturbations, $\Delta x$ and $I$ are responses.

推导（Derivation）

Total Hamiltonian,

$$H(t) = H_0 + H'(t)$$

where $H'(t)$ is the external perturbation.

$$H'(t) = -f(t) B$$

$B$ is operator, $f(t)$ is a c-number function, called generalized force（广义力）.

Such as:

$$- \hat S \cdot H(t)$$

(spin coupled to external magnetic field)

and,

$$- \frac{1}{c} \hat j \cdot A(t)$$

(current coupled to the vector potential)

密度算符（Density operator）

$$\rho = \sum\limits_n \left| n \right\rangle w_n \left\langle n \right|$$

fulfills the Quantum Liouville equation,

$$i \frac{\partial }{\partial t} \rho = \left[ H, \rho \right]$$

Suppose $\rho_0$ is the density operator in equilibrium,

$$i \frac{\partial}{\partial t} \rho_0 = \left[ H_0, \rho_0 \right] = 0$$

that means, in equilibrium, density operator doesn't change with time.

相互作用绘景（Interaction Picture）

$$\rho_I(t) = e^{i H_0 t} \rho e^{-i H_0 t}$$

$$H'_I(t) = - B_I f(t) = - e^{i H_0 t}B e^{-iH_0 t} f(t)$$

Quantum Liouville equation in IP,

$$i \frac{\partial }{\partial t} \rho_I = \left[ H_0 + H', \rho_I \right] = \left[ H', \rho_I \right] = - \left[ B_I, \rho_I \right] f(t)$$

Suppose, $t \to - \infty$, $\rho_I \to \rho_0$, then add on $H'$ adiabatically.

$$\rho_I(t) = \rho_0 + \frac{1}{i} \int_{-\infty}^t dt' \left[ H'(t'), \rho_I(t') \right]$$

Observable $A$,

$$\left\langle A \right\rangle = \text{Tr} \rho A$$

$$\Delta A = \left\langle A \right\rangle - \left\langle A \right\rangle_0 = \text{Tr} \rho A - \text{Tr} \rho_0 A$$

consider,

$$\text{Tr} \left[ H', \rho \right] A = \text{Tr}H' \rho A - \text{Tr} \rho H' A = \text{Tr} \rho A H' - \text{Tr} \rho H' A = \left\langle \left[ A, H' \right] \right\rangle$$

We get,

$$\Delta A(t) = \frac{1}{i} \int_{-\infty}^t dt' \left\langle \left[ A(t), H'(t') \right] \right\rangle$$

First order perturbation,

$$\left\langle ... \right\rangle \to \left\langle ... \right\rangle_0$$

Then,

$$\Delta A(t) = - \frac{1}{i} \int_{-\infty}^t dt' \left\langle \left[ A(t), B(t') \right] \right\rangle_0 f(t')$$

i.e.,

$$\Delta A(t) = - \frac{1}{i} \int_{-\infty}^{\infty} dt' \theta(t-t') \left\langle \left[ A(t), B(t') \right] \right\rangle_0 f(t')$$

Let's introduce the retarded Green's function of operators $A$ and $B$.

$$\chi^R_{AB}(t,t') = \frac{1}{i} \left\langle \left[ A(t), B(t') \right] \right\rangle_0 \theta(t-t')$$

also, the advanced Green's function,

$$\chi^A_{AB}(t,t') = - \frac{1}{i} \left\langle \left[ A(t), B(t') \right] \right\rangle_0 \theta(t'-t)$$

Use $A(t) = e^{i H_0} A e^{- i H_0}$, we can prove,

$$\left\langle \left[ A(t), B(t') \right] \right\rangle = \left\langle \left[ A(t - t'), B \right] \right\rangle$$

Thus, the Kubo formula,

$$\Delta A(t ) = - \int_{-\infty}^\infty dt' \chi^R_{AB}(t-t')f(t')$$

In Convolution（卷积） form,

$$\Delta A = - \chi^R_{AB} \ast f$$

FT,

$$\Delta A(\omega) = - \chi_{AB}^R(\omega) f(\omega)$$

Specific examples,

• electric conductivity 电导率, $\bf{j} (\omega) = \sigma(\omega) \bf{E}(\omega)$

• magnetic susceptibility 磁化率, $\bf{m}(\omega) = \chi(\omega) \bf{H}(\omega)$

结论（Conclusion）

To study electric conductivity, we should calculate the current-current correlation in equilibrium, i.e.,

$$\left\langle \hat j(t) \hat j(t') \right\rangle_0$$

Which means, the linear response of the system to the external electric field, the nonequilibrium current ($\Delta \hat j(t)$), turns out to be determined by the equilibrium correlations of the current itself ($\left\langle \hat j(t) \hat j(t') \right\rangle_0$).

The external field (耗散), in a sense, simply reveals these equilibrium fluctuations (涨落).

涨落-耗散定理（Fluctuation-Dissipation Theorem）

Autocovariation

Autocorrelation function 自相关函数

$$K_A (t,t') = \frac{1}{2}\left\langle \left\{ A(t), A(t') \right\} \right\rangle$$

$t' \to 0$, $t \to t$

$$K_A(t) = \frac{1}{2}\left\langle \left\{ A(t), A(0) \right\} \right\rangle$$

Autocovariation function,

$$K_{\Delta A}(t) = \frac{1}{2} \left\langle \left\{ \Delta A(t), \Delta A(0) \right\} \right\rangle$$

$$\Delta A(t) = A(t)- \left\langle A \right\rangle$$

$$\Delta A(0) = A(0)- \left\langle A \right\rangle$$

use,

$$\left\langle A(t) \right\rangle = \left\langle A(0) \right\rangle = \left\langle A \right\rangle$$

we get,

$$K_{\Delta A}(t) = K_A(t) - \left\langle A \right\rangle^2$$

涨落的谱密度（Spectral density of fluctuation）

FT of $K_A(t)$,

$$K_A(\omega) = \int_{-\infty}^{\infty} dt e^{i \omega t } K_A (t)$$

$K_A(\omega)$ is often denoted as $S(\omega)$.

Kubo-Martin-Schwinger identity:

$$\left\langle A(t)B(0) \right\rangle = e^{\beta \Omega} \text{Tr} \left\{ e^{-\beta H} e^{i Ht}A(0)e^{-iHt}B(0) \right\}$$

$$= e^{\beta \Omega} \text{Tr} \left\{ e^{-\beta H} B e^{-\beta H} e^{i Ht} A e^{-i Ht} e^{\beta H} \right\} = \left\langle B(0) A(t + i \beta) \right\rangle$$

Then,

$$\left\langle A(t)A(0) \right\rangle = \left\langle A(0)A(t + i \beta) \right\rangle$$

$$K_A(\omega) = \frac{1}{2} \int_{-\infty}^\infty dt e^{i \omega t } \left\langle A(t)A(0) + A(0)A(t) \right\rangle$$

$$= \frac{1}{2} \int_{-\infty}^\infty dt e^{i \omega t } \left\langle A(0)A(t) + A(0)A(t+i \beta) \right\rangle$$

1st term,

$$\left\langle A(0)A(t) \right\rangle_\omega = \int_{-\infty}^\infty dt e^{i \omega t} \left\langle A(0)A(t) \right\rangle$$

2nd term,

$$\left\langle A(0)A(t+i \beta) \right\rangle_\omega = \int_{-\infty}^\infty dt e^{i \omega t} \left\langle A(0)A(t+i \beta) \right\rangle$$

change of variable,

$$t + i \beta \to t$$

$$t \to t - i \beta$$

$$dt \to dt$$

$$\left\langle A(0)A(t+i \beta) \right\rangle_\omega = \int_{-\infty}^\infty dt e^{i \omega t} e^{ \beta \omega } \left\langle A(0)A(t) \right\rangle = e^{ \beta \omega } \int_{-\infty}^\infty dt e^{i \omega t} \left\langle A(0)A(t) \right\rangle$$

We get,

$$K_A(\omega) = \frac{1+ e^{\beta \omega}}{2} \int_{-\infty}^\infty dt e^{i \omega t} \left\langle A(0)A(t) \right\rangle$$

Likewise, we calculate the formula for commutator $\left[ A(t), A(0) \right]$,

$$\left\langle \left[ A(t), A(0) \right] \right\rangle_\omega = \int_{-\infty}^\infty dt e^{i \omega t} \left\langle \left[ A(t), A(0) \right] \right\rangle$$

$$= \left( e^{\beta \omega } - 1 \right) \int_{-\infty}^\infty dt e^{i \omega t} \left\langle A(0)A(t) \right\rangle$$

Then,

$$K_A(\omega) = \frac{1}{2} \coth \frac{\beta \omega}{2} \left\langle \left[ A(t), A(0) \right] \right\rangle_\omega$$

Recall that,

$$\chi_{AA}^R(t) = \frac{1}{i} \left\langle [ A(t), A(0) ] \right\rangle \theta(t)$$

$$\chi_{AA}^A(t) = - \frac{1}{i} \left\langle [ A(t), A(0) ] \right\rangle \theta(- t)$$

Then,

$$\left\langle \left[ A(t), A(0) \right] \right\rangle_\omega = \int_{-\infty}^0 dt e^{i \omega t} \left\langle \left[ A(t), A(0) \right] \right\rangle + \int_0^\infty dt e^{i \omega t} \left\langle \left[ A(t), A(0) \right] \right\rangle$$

$$= \int_{-\infty}^\infty dt e^{i \omega t} \left\langle \left[ A(t), A(0) \right] \right\rangle \theta(-t) + \int_{-\infty }^\infty dt e^{i \omega t} \left\langle \left[ A(t), A(0) \right] \right\rangle \theta(t)$$

$$= i \int_{-\infty}^\infty dt e^{i \omega t} \left( \chi_{AA}^R(t) - \chi^A_{AA}(t) \right) = i \left(\chi_{AA}^R (\omega) - \chi_{AA}^A(\omega) \right) = - 2 \Im \chi^R_{AA}(\omega)$$

Finally,

$$S(\omega) = K_A(\omega) = - \coth \frac{\beta\omega}{2} \Im \chi_{AA}^R(\omega)$$

It is the imaginary part of susceptibility that determines the energy dissipation rate in the system, hence the name - Fluctuation-Dissipation theorem.

References

• Alexandre M. Zagoskin, Quantum Theory of Many-Body Systems, $\S$3.3;

• Wiki: Fluctuation-dissipation theorem

@季燕江